advanced functions chapter 6 solutions

MHR • Advanced Functions 12 Solutions 585 585

Chapter 6 Exponential and Logarithmic Functions Chapter 6 Prerequisite Skills Chapter 6 Prerequisite Skills Question 1 Page 308 a)

b) i) Domain:

x !R{ }

ii) Range:

y !R, y > 0{ }

iii) y = 0 Chapter 6 Prerequisite Skills Question 2 Page 308 a) P(0) = 300(2)0 = 300(1) = 300 b) P(3) = 300(2)3 = 300(8) = 2400 c) i) ii)

t &= 2.74 days approximately 38 bacteria Chapter 6 Prerequisite Skills Question 3 Page 308 a) x7 b) m3 c) k6 d) –8x12

e)

!2b

a f)

2

x2

g)

1

u


Chapter 6 Prerequisite Skills Question 4 Page 308

a) 22= 4 b)

103

102= 10

c)

�

1

3

!

" #

$

% &

6

1

3

9!

"

# #

$

%

& &

=1

3

!

" #

$

% &

'3

= 33

= 27

d)

�

36!3

5

37

+1=3

11

37

+1

= 34

+1

= 81+1

= 82

Chapter 6 Prerequisite Skills Question 5 Page 308 a)

b) Domain:

x !R, x " 0{ } , Range:

y !R, y " 0{ }

c) d)

e) Yes. f –1 is a function since it passes the vertical line test. f) Domain:

x !R, x " 0{ } , Range:

y !R, y " 0{ }


Chapter 6 Prerequisite Skills Question 6 08 a)

b) Domain:

x !R{ } , Range:

y !R, y " 4{ }

c)

d)

e) No. f –1 is not a function since it does not pass the vertical line test. f) Domain:

x !R, x " 4{ } , Range:

y !R{ }

Chapter 6 Prerequisite Skills Question 7 Page 308 Yes. Each curve is a reflection of the other in the line y = x. Chapter 6 Prerequisite Skills Question 8 Page 309 a) Translation of 3 units to the right and 1 unit up. b) Reflection in the x-axis and a vertical stretch of factor 2.


Chapter 6 Prerequisite Skills Question 9 Page 309 a) A vertical stretch of factor 3 and vertical translation of 4 units down. b) The period is doubled followed by a reflection in the y-axis. Chapter 6 Prerequisite Skills Question 10 Page 309

Apply a horizontal translation of 2 units to the right and a vertical translation of 7 units down. Chapter 6 Prerequisite Skills Question 11 Page 309

Apply a vertical stretch by a factor of 2 and a reflection in the x-axis. Chapter 6 Prerequisite Skills Question 12 Page 309

Apply a vertical stretch by a factor of 3 and a reflection in the line y = x.


Chapter 6 Section 1 The Exponential Function and Its Inverse Chapter 6 Section 1 Question 1 Page 318 C, D First Differences A B

C D

If you look at the first differences, y is increasing at a rate proportional to the functions for C and D so they are exponential.


Chapter 6 Section 1 Question 2 Page 318 C:

y = bx

1= b0 This statement is true for any value of b.

3= b1

b = 3 The only valid value for b is 3.

Check:

9 = 32

Since b = 3, y = 3x.

D:

y = bx

1= b0 This statement is true for any value of b.

1

3= b1

b =1

3 The only valid value for b is 3.

Check:

9 =1

3

!"#

$%&

'2

Since b =1

3, y =

1

3

!"#

$%&

x

.


Chapter 6 Section 1 Question 3 Page 318 a)

b) i)

m12=

1.52!1.5

1

2 !1

= 0.75

ii)

m23=

1.53!1.5

2

3! 2

&= 1.13

iii)

m34=

1.54!1.5

3

4 ! 3

&= 1.69

iv)

m45=

1.55!1.5

4

5! 4

&= 2.53

c)

m1=

1.51!1.5

0.999

1! 0.999

&= 0.6

m2=

1.52!1.5

1.999

2 !1.999

&= 0.9

m3=

1.53!1.5

2.999

3! 2.999

&= 1.4

m4=

1.54!1.5

3.999

4 ! 3.999

&= 2.1

m5=

1.55!1.5

4.999

5! 4.999

&= 3.1

d) The rates are increasing.



b) i)

�

m!3!2

=0.5

!2!0.5

!3

!2 + 3

= !4

ii)

�

m!2!1

=0.5

!1!0.5

!2

!1+ 2

= !2

iii)

�

m!10

=0.5

0!0.5

!1

0 +1

= !1

iv)

�

m01

=0.5

1!0.5

0

1!0

= !0.5

c)

m!3=

0.5!3! 0.5

!2.999

!3+ 2.999

&= !5.5

m!2=

0.5!2! 0.5

!1.999

!2 +1.999

&= !2.8

m!1=

0.5!1! 0.5

!0.999

!1+ 0.999

&= !1.4

m0=

0.50! 0.5

0.0001

0 ! 0.0001

&= !0.7

m1=

0.51! 0.5

0.999

1! 0.999

&= !0.35

d) The rates are increasing.



b) y = 3x c), d)


b)

y =1

3

!"#

$%&

x

c), d)


Chapter 6 Section 1 Question 7 Page 319 a) iii) b) i) c) ii) d) iv) Chapter 6 Section 1 Question 8 Page 320

a) Graph is the inverse of 7d): y =

1

5

!"#

$%&

x

b) Graph is the inverse of 7b): y = 5x

c) Graph is the inverse of 7c): y =

1

2

!"#

$%&

x

d) Graph is the inverse of 7a): y = 2x



b)

c) All three functions have the same domain and are increasing. d) h(x) = 3x is different than the other two functions for range, x-intercept, y-intercept, y = 0 asymptote and positive/negative intervals. e) For f(x) = 3x, the instantaneous rate of change is constant. For g(x) = x3, the instantaneous rate of change is decreasing then increasing. For h(x) = 3x, the instantaneous rate of change is increasing.


Chapter 6 Section 1 Question 10 Page 320 a) i) N(0) = 10(2)0 ii) N(1) = 10(2)1 = 10 = 20 iii) N(2) = 10(2)2 iv) N(3) = 10(2)3 = 40 = 80 b) Yes, the function appears to be exponential since the rate of change is increasing.

c)

m12=

10(2)2!10(2)1

2 !1

= 20

d) i)

m1=

10(2)1!10(2)0.999

1! 0.999

&= 13.9

ii)

m2=

10(2)2!10(2)1.999

2 !1.999

&= 27.7

e) Answers may vary. A sample solution is shown. c) is an average rate of change between two

points. d) is an instantaneous rate at one point. If you average the two instantaneous rates in d) (20.8), it is close to the value in c) (20).

Chapter 6 Section 1 Question 11 Page 320 a) b), c)


Chapter 6 Section 1 Question 12 Page 320 a) Domain:

x !R{ } , Range:

y !R, y > 0{ }

b) No x-intercept. c) The y-intercept is 1. d) The function is never negative. The function is positive for all intervals. e) The function is increasing for all intervals. f) The horizontal asymptote is y = 0. Chapter 6 Section 1 Question 13 Page 321 a) Domain:

x !R, x > 0{ } , Range:

y !R{ }

b) The x-intercept is 1. c) No y-intercept. d) For 0 < x < 1, the function is negative. For x > 1, the function is positive. e) The function is increasing for x > 0. f) The vertical asymptote is x = 0. Chapter 6 Section 1 Question 14 Page 321 a) b), c)



x !R{ } , Range:

y !R, y > 0{ }

b) No x-intercept. c) The y-intercept is 1. d) The function is positive for all intervals. e) The function is decreasing for all intervals. f) The horizontal asymptote is y = 0. Chapter 6 Section 1 Question 16 Page 321 a) Domain:

x !R, x > 0{ } , Range:

y !R{ }

b) The x-intercept is 1. c) No y-intercept. d) For 0 < x < 1, the function is positive. For x > 1, the function is negative. e) The function is decreasing for x > 0. f) The vertical asymptote is x = 0. Chapter 6 Section 1 Question 17 Page 321 a) The domain, range, y-intercept, and the horizontal asymptote are the same. They are positive

for the same interval.

b) y = 4x increases and

y =1

2

!"#

$%&

x

decreases.

Chapter 6 Section 1 Question 18 Page 321 a) The domain, range, x-intercept, and vertical asymptote are the same. b) f –1 and g–1 are positive and negative for different intervals. f –1 is increasing for all intervals

and g–1 is decreasing for all intervals.


Chapter 6 Section 1 Question 19 Page 321 a) b), c)

d) y = 6x Chapter 6 Section 1 Question 20 Page 321

y =1

10

!"#

$%&

x


x y 0 1 1 –2 2 4 3 –8 4 16

b)

c) No, the points do not form a smooth curve, they form a v-shape.


d) i) ii)

The values are undefined because they are not real (square root of a negative value). e) Answers may vary. A sample solution is shown. Exponential functions are defined only for functions with positive bases since they are real. Exponential functions with negative bases are not real.

Chapter 6 Section 1 Question 22 Page 321

a)

m12=

1.42!1.4

1

2 !1

= 0.56

m34=

1.44!1.4

3

4 ! 3

&= 1.10

The ship’s average velocity between 1 s and 2 s is 56 km/s. The ship’s average velocity between 3 s and 4 s is approximately 110 km/s.

b)

m3=

1.43!1.4

2.999

3! 2.999

&= 0.92

m4=

1.44!1.4

3.999

4 ! 3.999

&= 1.29

The ship’s instantaneous velocity is 92 km/s at 3 s and 129 km/s at 4 s. Chapter 6 Section 1 Question 23 Page 322 a) For 0 < b < 1, f and f –1 have equal x- and y-coordinates at the point where they intersect the

line y = x. b) Yes; when b > 1, the graphs do not intersect the line y = x.


Chapter 6 Section 1 Question 24 Page 322 a), b)

c) i) f is a function for b > 0. ii) f is undefined for b < 0. d) i) f –1 is a function for b > 0. ii) f –1 is undefined for b < 0. e) Yes, it seems that if f is a function then f –1 is a function. If f is undefined, f –1 is undefined.


Chapter 6 Section 2 Logarithms Chapter 6 Section 2 Question 1 Page 328

a) log

464 = 3 b)

log

2128 = 7 c)

log5

1

25

!"#

$%&= '2 d)

log1

2

0.25 = 2

e) log

6y = x f)

log

10100 000 = 5 g)

log3

1

27

!"#

$%&= '3 h)

log

bv = u


a)

y = log2

64

2 y= 64

2 y= 26

y = 6

b)

y = log3

27

3y= 27

3y= 33

y = 3

c)

y = log2

1

4

!"#

$%&

2 y=

1

4

2 y=

1

22

2 y= 2'2

y = '2

d)

y = log4

1

64

!"#

$%&

4 y=

1

64

4 y=

1

43

4 y= 4'3

y = '3

e)

y = log5125

5y= 125

5y= 53

y = 3

f)

y = log21024

2 y= 1024

2 y= 210

y = 10

g)

y = log6

363

6 y= 363

6 y= 66

y = 6

h)

y = log381

3y= 81

3y= 34

y = 4


a)

y = log10

1000

10 y= 1000

10 y= 103

y = 3

b)

y = log10

1

10

!"#

$%&

10 y=

1

10

10 y= 10'1

y = '1

c)

y = log10

1

10 y= 1

10 y= 100

y = 0

d)

y = log10

0.001

10 y= 0.001

10 y= 10!3

y = !3


e)

y = log10

10!4

10 y= 10!4

y = !4

f)

y = log10

1 000 000

10 y= 1 000 000

10 y= 106

y = 6

g)

y = log10

1

100

!"#

$%&

10 y=

1

100

10 y= 10'2

y = '2

h)

y = log10

10 000

10 y= 10 000

10 y= 104

y = 4

Chapter 6 Section 2 Question 4 Page 328 a) 72 = 49 b) 25 = 32 c) 104 = 10 000 d) bw = z

e) 23 = 8 f) 54 = 625 g)

10!2=

1

100 h) 72y = x

Chapter 6 Section 2 Question 5 Page 328 a) y = 2x x = 2y or y = log2 x

b) y = 4x x = 4y or y = log4 x



2

x= 6; y

1= 2

x, y

2= 6 b)

4

x= 180; y

1= 4

x, y

2= 180

log

26 &= 2.6

log

4180 &= 3.7

c)

3

x= 900; y

1= 3

x, y

2= 900 d)

9

x= 0.035; y

1= 9

x, y

2= 0.035

log

3900 &= 6.2

log

90.035 &= !1.5

Chapter 6 Section 2 Question 7 Page 328 Answers may vary. A sample solution is shown. d)

9x= 0.035

9x! 0.035 = 0

Solve by graphing y = 9

x! 0.035 and finding the zero.

log

90.035 &= !1.5


Chapter 6 Section 2 Question 8 Page 328 Evaluate each expression using the log function on a calculator. An example is shown for part a). a) 2.63

b) –4.43 c) 0.95 d) –0.70 e) 1.23 f) 2.00 g) 2.26 h) 3.00 Chapter 6 Section 2 Question 9 Page 329 a) y = 10x

b)


a)

y = log33

3y= 3

3y= 31

y = 1

b)

y = log2

2

2 y= 2

2 y= 21

y = 1

c)

y = log12

12

12 y= 12

12 y= 121

y = 1

d)

y = log1

2

1

2

1

2

y

=1

2

1

2

y

=1

2

1

y = 1


Chapter 6 Section 2 Question 11 Page 329 a) logx x = 1 for x > 0, x ≠ 1 b) Answers may vary. A sample solution is shown.

log0.1

0.1= 1

log1

3

1

3= 1

log11

11= 1

c)

log

xx = 1, x > 0, x ! 1

L.S. R.S.

y = logx

x y = 1

xy= x

xy= x

1

y = 1

Since L.S. = R.S., log

xx = 1, x > 0, x ! 1 is true.

Chapter 6 Section 2 Question 12 Page 329 a) Answers may vary. A sample solution is shown. y = log x

y = 10

x

The logarithmic function has decreasing slope (the rate of change is decreasing) and the exponential function has increasing slope (the rate of change is increasing).

b) Answers may vary. A sample solution is shown.

The average rate of change of a logarithmic function is the reciprocal of the average rate of change of its inverse, which is an exponential function.

log(100)! log(10)

100 !10=

1

90

102!10

1

2 !1= 90



It takes approximately 6.3 days for the number of visitors to the Web site to reach 1000. b)

It takes approximately 12.6 days for the number of visitors to the Web site to reach 1 000 000. Chapter 6 Section 2 Question 14 Page 329 a)

The car is approximately 66.5 m away.


b)

No, if the headlight intensity doubles the car is approximately 16.2 m away, which is closer than half.

c) Answers may vary. A sample solution is shown. Drive slower. Chapter 6 Section 2 Question 15 Page 329 a)

The thickness of the hull needs to be at least 5.72 cm. b)

The thickness of the hull needs to be at least 7.15 cm, so it needs to be increased by approximately 1.43 cm.


Chapter 6 Section 2 Question 16 Page 329 Solutions to Achievement Check questions are provided in the Teacher’s Resource. Chapter 6 Section 2 Question 17 Page 330 Answers may vary. A sample solution is shown.

a) Approximately 3 integer values can be viewed at one time. b) It is difficult to view a broad range of this function since y is an integer when x is a power of

10. For example: x y 10 1

100 2 1000 3

Chapter 6 Section 2 Question 18 Page 330 a) Answers may vary. A sample solution is shown.

x y 100 0 101 1 102 2 103 3 104 4 105 5 106 6


b)

The graph is linear. The semi-log grid has turned each power into the exponent value. c) An advantage is to plot a greater range of values. Chapter 6 Section 2 Question 19 Page 330 a)

1 0 10 1

100 2 1000 3

10000 4 100000 5

1000000 6 b)

The graph is a curve with positive, decreasing slope.


c) 1 0 0

10 1 1 100 2 2

1000 3 3 10000 4 4

100000 5 5 1000000 6 6

d)

The graph is linear with positive, constant slope. e) Answers may vary. A sample solution is shown. You can graph a greater range of values. Chapter 6 Section 2 Question 20 Page 330

�

D;10

20061+10

2( )2!10

2007=

1+100

2!10

=101

20

= 5.05


Chapter 6 Section 2 Question 21 Page 330 Method 1:

m!1

m= 1 n !

1

n= !1

m2! m!1= 0 n2

+ n !1= 0

m =1± (!1)2

! 4(1)(!1)

2(1) n =

!1± (1)2! 4(1)(!1)

2(1)

m =1+ 5

2 n =

!1+ 5

2 m, n are positive

m+ n =1+ 5

2+!1+ 5

2

=2 5

2

= 5

Method 2:

m!1

m= 1

m2! m!1= 0

m =1± (!1)2

! 4(1)(!1)

2(1)

m =1+ 5

2

m+1

m=

1+ 5

2+

2

1+ 5

=(1+ 5)(1+ 5)+ 2(2)

2(1+ 5)

=1+ 2 5 + 5+ 4

2(1+ 5)

=5+ 5

1+ 5

=5+ 5

1+ 5"

1! 5

1! 5

=5! 4 5 ! 5

1! 5

=!4 5

!4

= 5


Chapter 6 Section 3 Transformations of Logarithmic Functions Chapter 6 Section 3 Question 1 Page 338 a) iv; translated right 3 units b) ii; translated down 3 units c) i; translated left 3 units d) iii; translated up 3 units Chapter 6 Section 3 Question 2 Page 338 a) translate right 2 units b) translate left 5 units and down 4 units c) translate up 1 unit d) translate left 4 units and down 6 units Chapter 6 Section 3 Question 3 Page 338 a)

Domain:

x !R, x > 0{ } ; Range:

y !R{ }

x-intercept, let y = 0 y-intercept, let x = 0 0 = log x + 2 y = log 0 + 2 –2 = log x 10–2 = x x = 0.01 The x-intercept is 0.01. Since log 0 is undefined, there is no y-intercept. The vertical asymptote is x = 0.


b)

Domain:

x !R, x > 3{ } ; Range:

y !R{ }

x-intercept, let y = 0 y-intercept, let x = 0 0 = log(x – 3) y = log(0 – 3) 100 = x – 3 y = log –3 x = 1 + 3 x = 4 The x-intercept is 4. Since log –3 is undefined, there is no y-intercept. The vertical asymptote occurs when x – 3 = 0. The vertical asymptote is x = 3. c)

Domain:

x !R, x > 3{ } ; Range:

y !R{ }

x-intercept, let y = 0 y-intercept, let x = 0 0 = log(x – 3) + 4 y = log(0 – 3) + 4 –4 = log(x – 3) y = log –3 + 4 10–4 = x – 3 x = 0.0001 + 3 x = 3.0001 The x-intercept is 3.0001. Since log –3 is undefined, there is no y-intercept. The vertical asymptote occurs when x – 3 = 0. The vertical asymptote is x = 3.


d)

Domain:

x !R, x > "5{ } ; Range:

y !R{ }

x-intercept, let y = 0 y-intercept, let x = 0 0 = log(x + 5) – 1 y = log(0 + 5) – 1 1 = log(x + 5) y = log 5 – 1 101 = x + 5 y &= –0.301 x = 10 – 5 x = 5 The x-intercept is 5. The y-intercept is approximately –0.301. The vertical asymptote occurs when x + 5 = 0. The vertical asymptote is x = –5. Chapter 6 Section 3 Question 4 Page 338 a) y = 2log x; vertical stretch by a factor of 2

b) y = log(2x); horizontal compression by a factor of

1

2

c) y = log

1

2x

!"#

$%&

; horizontal stretch by a factor of 2

d) y =

1

2log x; vertical compression by a factor of

1

2


a) vertical compression by a factor of

1

2

b) horizontal compression by a factor of

1

5 and a reflection in the y-axis

c) horizontal stretch by a factor of 2 and a reflection in the y-axis d) vertical stretch by a factor of 5 and a reflection in the x-axis


Chapter 6 Section 3 Question 6 Page 339 a) b)

c) d)



Chapter 6 Section 3 Question 8 9 For y = –log(x – 4) For y = log(–x) – 3 a) Domain:

x !R, x > 4{ } a) Domain:

x !R, x < 0{ }

b) Range:

y !R{ } b) Range:

y !R{ }

c) x = 4 c) x = 0 Chapter 6 Section 3 Question 9 Page 339 a)

b) Range:

y !R, 2 " y " 6{ }


Chapter 6 Section 3 Question 10 Page 339 Answers may vary. A sample solution is shown. The domain and range remain the same under a vertical reflection. y = log x and y = –log x y = log(x – 2) and y = –log(x – 2)

Domain:

x !R, x > 0{ } ; Range:

y !R{ } Domain:

x !R, x > 2{ } ; Range:

y !R{ }

y = 3 log x and y = –3 log x y = log 4x and y = –log 4x

Domain:

x !R, x > 0{ } ; Range:

y !R{ } Domain:

x !R, x > 0{ } ; Range:

y !R{ }

y = log x – 2 and y = –log x – 2

Domain:

x !R, x > 0{ } ; Range:

y !R{ }


Chapter 6 Section 3 Question 11 Page 339 Answers may vary. A sample solution is shown. No, the same thing does not happen under a horizontal reflection. The domains are different, but the ranges are the same. y = log x and y = log(–x)

y = log x; Domain:

x !R, x > 0{ } ; Range:

y !R{ }

y = log(–x); Domain:

x !R, x < 0{ } ; Range:

y !R{ } y = 2 log x and y = 2 log(–x)

y = 2 log x; Domain:

x !R, x > 0{ } ; Range:

y !R{ }

y = 2 log(–x); Domain:

x !R, x < 0{ } ; Range:

y !R{ } y = log x – 2 and y = log(–x) – 2

y = log x – 2; Domain:

x !R, x > 0{ } ; Range:

y !R{ }

y = log(–x) – 2; Domain:

x !R, x < 0{ } ; Range:

y !R{ }


Chapter 6 Section 3 Question 12 Page 340 Answers may vary. A sample solution is shown. The domains are different, but the ranges are the same. y = log x and y = –log(–x)

y = log x; Domain:

x !R, x > 0{ } ; Range:

y !R{ }

y = –log(–x); Domain:

x !R, x < 0{ } ; Range:

y !R{ } y = 2 log(x) – 1 and y = –2 log(–x) – 1

y = 2 log(x) – 1; Domain:

x !R, x > 0{ } ; Range:

y !R{ }

y = –2 log(–x) – 1; Domain:

x !R, x < 0{ } ; Range:

y !R{ }


Chapter 6 Section 3 Questions 13 and 14 Page 340 a) b)

vertical stretch by a factor of 2 reflection in the y-axis

horizontal compression by a factor of

1

3

translation 4 units to the left. translation 2 units down c) d)

vertical stretch by a factor of 5 vertical stretch by a factor of 2

horizontal compression by a factor of

1

2 reflection in the x-axis

translation 2 units to the right horizontal stretch by a factor of 2 translation 6 units down translation 6 units to the left translation 3 units down



b)

Vo = 6 Vo &= 6.3

For an input voltage of 10 V, the output is 6 V. For an input voltage of 20 V, the output is approximately 6.3 V.

c)

Vi = 1020 The input voltage is 1020 V when the output is 25 V. d) The domain is the input voltage, {Vi ∈ R, Vi > 0}; the range is the output voltage, Vo, which

can be any real number: {Vo ∈ R}. Chapter 6 Section 3 Question 16 Page 340 Solutions to Achievement Check questions are provided in the Teacher’s Resource.



c)

d) i) Domain:

x !R{ }

ii) Range:

y !R, y > 0{ } iii) y = 0 e) y = 102x + 8 Chapter 6 Section 3 Question 18 Page 340 a) b)



t !R, 2n" < t < (2n +1)" , n !Z{ } ; Range:

V

o!R, V

o" 5{ }

Domain: The range of y = sin t is

y !R, "1# y #1{ } , but for y = log x,

x !R, x > 0{ } , so sin t > 0. t > 0 or t > 2π or t > 4π or t > 2nπ, n ∈ Z For 0 < sin t ≤ 1, look at the graph y = sin t.

We can see from the graph that 0 < sin t ≤ 1 when 0 < t ≤ π or –2π < t ≤ –π or 2nπ < t ≤ (2n + 1)π, n ∈ Z. The range is found by substituting the largest value for sin t = 1, Vo = 5. b)

c) Answers may vary. A sample solution is shown. Because the pattern is a beat, like a pulse. Chapter 6 Section 3 Question 20 Page 340

x = 64


Chapter 6 Section 4 Power Law of Logarithms Chapter 6 Section 4 Question 1 Page 347 a)

log2(24 )3

= log2

212

= 12

b) log

464 = 3

c)

log 102( )!4

= log10!8

= !8

d)

1

2log(10!1) =

1

2(!1)

= !1

2


a)

log2(23)

1

2 =1

2log

223

=1

2(3)

=3

2

b)

log3(35)

1

2 =1

2(5)

=5

2

c)

log3

(34 )

1

3!

"#

$

%&

6

= log338

= 8

d)

log4

(42 )

1

5!

"#

$

%&

15

= log4

46

= 6


a)

log10 = log 4t

1= t log 4

1

log 4= t

t &= 1.66

b)

log5t= log 250

t log5 = log 250

t =log 250

log5

t &= 3.43

c)

log 2 = log1.08t

log 2 = t log1.08

log 2

log1.08= t

t &= 9.01

d)

500

100= 1.06t

log5 = log1.06t

log5 = t log1.06

log5

log1.06= t

t &= 27.62


Chapter 6 Section 4 Question 4 Page 347 a) i)

A(0) = 500(1.07)0

= 500

The value of the investment is initially $500. ii)

A(2) = 500(1.07)2

= 572.45

The value of the investment after 2 years is $572.45. iii)

A(4) = 500(1.07)4

&= 655.40

The value of the investment after 4 years is approximately $655.40 b) i)

2 = 1.07t

log 2 = t log1.07

log 2

log1.07= t

t &= 10.2

It will take approximately 10.2 years for the investment to double in value. ii)

3= 1.07t

log3= t log1.07

log3

log1.07= t

t &= 16.2

It will take approximately 16.2 years for the investment to triple in value. Chapter 6 Section 4 Question 5 Page 347

a)

log 23

log3&= 2.854 b)

log 20

log6&= 1.672

c)

log 2

log7&= 0.356 d)

!log 4

log12&= !0.558

e)

log30

log1

2

&= !4.907 f)

log8

log3

4

&= !7.228


a) log

58 b)

log

917 c)

log2

3

1

2 d)

log

x!1( )x +1( )





a)

2 = x log3

2

log3= x

x &= 4.192

b)

100

10= x log1000

10 = x(3)

x &= 3.333

c)

4 = xlog15

log3

!"#

$%&

x =4

log15

log3

!"#

$%&

x &= 1.623

d)

12

2= x

log3

log5

!"#

$%&

x =6

log3

log5

!"#

$%&

x &= 8.790


A(0) = 400(1.09)0

= 400

The initial value of the investment is $400, when t = 0. b)

2 = 1.09t

log 2 = t log1.09

t =log 2

log1.09

t &= 8

The investment will double in approximately 8 years.


log(mx) = m log x

log(mx) = log xm

mx = xm

xm! mx = 0

x(xm!1

! m) = 0

so x = 0, but x ≠ 0 since log 0 is undefined or xm – 1 = m, true for m = 1 log(mx) = m log x only when m = 1.



log(xn ) = (log x)n only when n = 1 and/or when x = 1.


a)

log285

= log2

23( )5

= log2

215

= 15

b)

log2

85= 5log

28

= 5log2

23

= 5(3)

= 15


c) Answers may vary. A sample solution is shown. I prefer the power law method since it involves simpler calculations. Chapter 6 Section 4 Question 14 Page 348 Answers may vary. A sample solution is shown.

log3

27

1

3 = log3

33( )1

3

= log331

= 1

log3

27

1

3 =1

3log

333( )

=1

3(3)

= 1


a)

t = log0.5

10

1000

!"#

$%&

t = log0.5

0.01

t &= 6.6

; t &= 6.6 h

The docking procedures should begin approximately 6.6 h after the reverse thrusters are first fired.

b) Domain:

d !R, 0 < d "1000{ } ; Range:

t !R, t " 0{ }

The domain represents the distance from the space station in kilometres. The range represents the time, in hours, required to reach a distance from the space station. Chapter 6 Section 4 Question 16 Page 348 Solutions to Achievement Check questions are provided in the Teacher’s Resource. Chapter 6 Section 4 Question 17 Page 348

Let w = logb

x

x = bw write in exponential form

logk

x = logk

bw take log

kof both sides

logk

x = w logk

b apply the power law of logarithms

w =log

kx

logk

b

logb

x =log

kx

logk

b express in terms of x



a)

log2

9

log2

3 b)

log2

25

log210


log

2210( )

64

= 640


A(t) = P(1+ r)i

�

r =0.035

4

= 0.008 75 interest rate each quarter

i = 4t number of pay periods each year

A(t) = P 1.008 75( )4t

b) i)

2 = (1.008 75)4t

log 2 = 4t log1.008 75

4t =log 2

log1.008 75

t =log 2

4 log1.008 75

t &= 19.9

It will take approximately 19.9 years for the investment to double. ii)

3= (1.008 75)4t

log3= 4t log1.008 75

4t =log3

log1.008 75

t =log3

4 log1.008 75

t &= 31.5

It will take approximately 31.5 years for the investment to triple.



f (x) =1

1! x

f (a) =1

1! a substitute a in for x

1

1! a= 2 given: f (a) = 2

1

2= 1! a

1

4= 1! a square both sides

a =3

4

f (1! a) = f1

4

"#$

%&'

=1

1!1

4

=1

3

4

=1

3

2

=2

3



Graph y = x x x and view the table for when y is an integer.

So e = 256 and f = 128. Check

�

f = 256 256 256

= 256 256!16

= 256 4096

= 256!64

= 16 384

= 128

e+ f = 256 +128

= 384


Chapter 6 Section 5 Making Connections: Logarithmic Scales in the Physical Sciences


pH = ! log H+"

#$%

= ! log(0.01)

= !(!2)

= 2

b)

pH = ! log H+"

#$%

= ! log(0.000 034)

&= !(!4.5)

&= 4.5

c)

pH = ! log H+"

#$%

= ! log(10!9 )

= !(!9)

= 9

d)

pH = ! log H+"

#$%

= ! log(1.5&10!10 )

&= !(!9.8)

&= 9.8

Chapter 6 Section 5 Question 2 Page 353 a) 10!pH

= H+"# $%

H+= 10

!11

b)

10!pH

= H+"

#$%

10!3= H

+"#

$%

H+= 0.001

c)

10!pH

= H+"

#$%

10!8.5

= H+"

#$%

H+

&= 3.2 &10!9

d)

10!pH

= H+"

#$%

10!4.4

= H+"

#$%

H+

&= 0.000 039 8


The pH scale varies over several powers of 10. b) Answers may vary. A sample solution is shown. To ensure that most pH measurements are positive. Chapter 6 Section 5 Question 4 Page 353 a)

pH = ! log H+"

#$%

= ! log(0.000 01)

= !(!5)

= 5

b)

pH = ! log H+"

#$%

= ! log(2.5&10!11)

&= !(!10.6)

&= 10.6



b)

The pH level of acetic acid is approximately 2.7. c)

pH = ! log H+"

#$%

= ! log(1.9 &10!3)

&= !(!2.7)

&= 2.7



a)

!2" !

1= 10 log

I2

I1

#

$%&

'(

80 ! 60 = 10 logI

2

I1

"

#$%

&'

20 = 10 logI

2

I1

"

#$%

&'

2 = logI

2

I1

"

#$%

&'

102=

I2

I1

100 =I

2

I1

The ratio of intensities is 100. A shout is 100 times more intense than a conversation.

b)

!2" !

1= 10 log

I2

I1

#

$%&

'(

80 ! 30 = 10 logI

2

I1

"

#$%

&'

50 = 10 logI

2

I1

"

#$%

&'

5 = logI

2

I1

"

#$%

&'

105=

I2

I1

100 000 =I

2

I1

The ratio of intensities is 100 000. A shout is 100 000 times more intense than a whisper.



!2" !

1= 10 log

I2

I1

#

$%&

'(

150 "110 = 10 logI

2

I1

#

$%&

'(

40 = 10 logI

2

I1

#

$%&

'(

4 = logI

2

I1

#

$%&

'(

104=

I2

I1

10 000 =I

2

I1

A rock concert speaker sounds 10 000 times as intense as a loud car stereo. Chapter 6 Section 5 Question 8 Page 354

!2" !

1= 10 log

I2

I1

#

$%&

'(

60 " !1= 10 log

30 000

1

#$%

&'(

60 " !1

&= 45

!1

&= 15

The decibel level of a pin drop is approximately 15 dB. Chapter 6 Section 5 Question 9 Page 354

M = logI

I0

!

"#$

%&

M = log 10 000( )= 4

The earthquake in North Bay, Ontario, measured 4 on the Richter scale.



a)

M = logI

I0

!

"#$

%&

2.3= logI

I0

!

"#$

%&

102.3=

I

I0

I

I0

&= 199.53

The earthquake in Welland, Ontario, was approximately 199.53 times as intense as a standard low-level earthquake.

b)

M = logI

I0

!

"#$

%&

1.1= logI

I0

!

"#$

%&

101.1=

I

I0

I

I0

&= 12.59

199.53

12.59&= 15.85

The Welland earthquake was approximately 15.85 times as intense as the St. Catharines earthquake.



M = logI

I0

!

"#$

%&

8.9 = logI

I0

!

"#$

%&

108.9=

I

I0

I

I0

&= 794 328 235

The most intense earthquakes are approximately 794 328 235 times as intense as standard, low-level earthquakes. Chapter 6 Section 5 Question 12 Page 354

a)

m2! m

1= log

b1

b2

"

#$%

&'

0.12 ! (!1.5) = logb

1

b2

"

#$%

&'

1.62 = logb

1

b2

"

#$%

&'

101.62=

b1

b2

b1

b2

&= 41.69

Sirius appears to be approximately 41.69 times brighter than Betelgeuse.

b)

m2! m

1= log

b1

b2

"

#$%

&'

(!1.5)! m1= log 1.3(1010( )

m1= !1.5! log 1.3(1010( )

m1

&= !11.61

The apparent magnitude of the Sun is approximately –11.61.


Chapter 6 Section 5 Question 13 Page 354 Solutions to Achievement Check questions are provided in the Teacher’s Resource. Chapter 6 Section 5 Question 14 Page 354 Answers may vary. A sample solution is shown. I think that astronomers would be more interested in the absolute magnitude since it takes distance into consideration. Chapter 6 Section 5 Question 15 Page 355 a) i)

M2! M

1= log

b1

b2

"

#$%

&'

0.4 ! (!1.8) = logb

1

b2

"

#$%

&'

2.2 = logb

1

b2

"

#$%

&'

b1

b2

= 102.2

b1

b2

&= 158.5

The brightest star is approximately 158.5 times brighter than the second brightest, in absolute terms.

ii)

M2! M

1= log

b1

b2

"

#$%

&'

3.2 ! (!1.8) = logb

1

b2

"

#$%

&'

5 = logb

1

b2

"

#$%

&'

b1

b2

= 105

b1

b2

= 100 000

The brightest star is 100 000 times brighter than the least brightest, in absolute terms. b) The closest star is Biffidus-V, next is Cheryl-XI, and farthest away is Roccolus-III.


Chapter 6 Section 5 Question 16 Page 355 Answers may vary. A sample solution is shown. There have been several major earthquakes around the Pacific Ocean. The largest known earthquakes are: • Chile, 1960 – magnitude 9.5 • Indonesia, 2004 – magnitude 9.3 (Boxing Day tsunami) • Russia, 1737 – estimated magnitude 9.3 • Alaska, 1964 – magnitude 9.2 • Russia, 1952 –magnitude 9.0 There have been minor earthquakes (tremors) where I live, but nothing above 4.0. Chapter 6 Section 5 Question 17 Page 355 Answers may vary. A sample solution is shown. There is a logarithmic equation relating the apparent magnitude (how bright we see a star), m, and the star’s distance from earth, d, to the absolute magnitude (how bright the star would appear from 10 parsecs, or about 32.6 light years, away), M:

M &= m ! 5 logd

3.26

"#$

%&'!1

(

)*

+

,-

Chapter 6 Section 5 Question 18 Page 355 Answers may vary. A sample solution is shown. The eye sees luminosity on a logarithmic scale, so people can look at one object 1 000 000 times brighter than another, but they do not appear that different.


Chapter 6 Section 5 Question 19 Page 355 B;

logsin x

cos x =1

2

log cos x( )log sin x( )

=1

2

2 log(cos x) = log(sin x)

log cos2x( ) = log sin x( )

cos2x = sin x

cos2x ! sin x = 0

(1! sin2x)! sin x = 0

! sin2x ! sin x +1= 0

sin x =1± (!1)2

! 4(!1)(1)

2(!1)

sin x =1± 5

!2

sin x =!1+ 5

2, since sin x cannot be less than !1



The base of the triangle

= s2+ s

2

= 2s2

= 2s

Area of triangle

=1

2bh

=1

22s( )h

Area of square = s2

Area square = Area triangle

s2=

1

22s( )h

2s2= 2sh

h =2s

2

2s

!2

2

divide the s 's

multiply the numerator and denominator by 2 to get the square root out of the denominator

h =2 2s

2

h = 2s

s


Chapter 6 Section 5 Question 21 Page 355 B;

a + b = 3

(a + b)2= 32

a2+ 2ab+ b

2= 9

a2+ b

2( ) + 2ab = 9

7 + 2ab = 9 substitute a2+ b

2= 7

2ab = 2

ab = 1

a =1

b

a2+ b

2( )2

= 72

a4+ 2a

2b

2+ b

4= 49

a4+ 2

1

b

!"#

$%&

2

b2+ b

4= 49

a4+ 2

b2

b2

!

"#

$

%& + b

4= 49

a4+ 2 + b

4= 49

a4+ b

4= 47

Chapter 6 Section 5 Question 22 Page 355 D;

logb(a

2 ) = 3

b3= a

2

b3( )

2

3 = a2( )

2

3

b2= a

4

3

loga

b2= log

aa

4

3

=4

3


Chapter 6 Review Chapter 6 Review Question 1 Page 356 a)

i) Domain:

x !R{ } ; Range:

y !R, y > 0{ }

ii) Let y = 0.

�

0 = 3x

log 0 = x log 3 log 0 is undefined

No x-intercept. iii) Let x = 0.

y = 30

y = 1

The y-intercept is 1. iv) positive for all intervals v) increasing for all intervals vi) The horizontal asymptote is y = 0. b), c)

Domain:

x !R, x > 0{ } ; Range:

y !R{ } ; x-intercept is 1; no y-intercept;

f –1 is positive for x > 1 and negative for 0 < x < 1; f –1 is increasing for all intervals; vertical asymptote is x = 0


Chapter 6 Review Question 2 Page 356 a)

i) Domain:

x !R{ } ; Range:

y !R, y > 0{ }

ii) Let y = 0.

�

0 =1

4

x

log 0 = x log1

4log 0 is undefined

No x-intercept. iii) Let x = 0.

y =1

4

0

y = 1

The y-intercept is 1. iv) positive for all intervals v) decreasing for all intervals vi) The horizontal asymptote is y = 0. b), c)

Domain:

x !R, x > 0{ } ; Range:

y !R{ } ; x-intercept is 1; no y-intercept;

g–1 is positive for 0 < x < 1 and negative for x > 1; g–1 is decreasing for all intervals; vertical asymptote is x = 0



log

464 = 3 b)

log

328 = x c)

log

6y = 3 d)

log

2512 = 9

Chapter 6 Review Question 4 Page 356 a) 27 = 128 b) bx = n c) 35 = 243 d) b19 = 4 Chapter 6 Review Question 5 Page 356

log

250 &= 5.6

log

232 = 5 and

log

264 = 6 , so

log

250 must be between 5 and 6. Find the approximate exponent

to which 2 must be raised to give 50.

Estimate Check Analysis 5.5

25.5

&= 45.3 Too low. Try a higher value.

5.7 2

5.7&= 52.0 High, but close.

5.6 2

5.6&= 48.5 This is a good estimate.

log

250 &= 5.6


log216 = log

224

= 4

b)

log381= log

334

= 4

c)

log4

1

16

!"#

$%&= log

4

1

42

!"#

$%&

= log4

4'2

= '2

d)

log0.000 001= log10!6

= !6


Chapter 6 Review Question 7 Page 356

log

xx = 1, x > 0, x ! 1

logx

x = y

xy= x

xy! x = 0

x(xy!1

!1) = 0

x = 0, not a solution, since log0

0 is not defined

xy!1

= 1

( y !1) log x = log1

( y !1) log x = 0

y !1= 0, y = 1

Therefore logx

x = 1 for x > 0, x " 1.


b) i) Domain:

x !R, x > 5{ }

ii) Range:

y !R{ }

iii) The vertical asymptote occurs when x – 5 = 0. The vertical asymptote is x = 5.



b) i) Domain:

x !R, x < 0{ }

ii) Range:

y !R{ }

iii) The vertical asymptote occurs when –x = 0. The vertical asymptote is x = 0. Chapter 6 Review Question 10 Page 356

Chapter 6 Review Question 11 Page 356 a) Answers may vary. A sample solution is shown. Horizontal stretch by a factor of 2. Translation 8 units left. Translation 3 units down. Reflection in the x-axis.

y =

! log1

2x + 4

"#$

%&'! 3 or

y = ! log1

2(x + 8)

"

#$

%

&' ! 3

b) Answers may vary. A sample solution is shown.


c) i) Domain:

x !R, x > "8{ }

ii) Range:

y !R{ }

iii) The vertical asymptote occurs when

1

2x + 4 = 0 . The vertical asymptote is x = –8.


log2

323= 3log

225( )

= 3(5)

= 15

b)

log1000!2= !2 log103

= !2 3( )= !6

c)

log0.001!1= ! log10!3

= ! !3( )= 3

d)

log1

4

1

16

!"#

$%&

4

= 4 log1

4

1

4

!"#

$%&

2

= 4(2)

= 8


a)

x =log17

log3

x &= 2.579

b)

x =log0.35

log 2

x &= !1.515

c)

x log 4 = log10

x =1

log 4

x &= 1.661

d)

80

100=

1

2

!"#

$%&

x

0.8 =1

2

!"#

$%&

x

log0.8 = x log1

2

x =log0.8

log1

2

x &= 0.322



0.75 = (0.8)x

log0.75 = x log0.8

x =log0.75

log0.8

x &= 1.3

The glass should be approximately 1.3 mm thick. Chapter 6 Review Question 15 Page 357 a) Draw a graph of the inverse function y = 3x, then reflect it in the line y = x.

b)


Chapter 6 Review Question 16 Page 357 Chemical A Chemical B

pH = ! log H+"

#$%

10!pH= H

+"#

$%

10!6= H

+

pH = ! log H+"

#$%

10!pH= H

+"#

$%

10!8= H

+

There is less hydronium ion concentration in Chemical B. Chapter 6 Review Question 17 Page 357 great light

M = logI

I0

!

"#$

%&

8 = logI

I0

!

"#$

%&

108=

I

I0

M = logI

I0

!

"#$

%&

4 = logI

I0

!

"#$

%&

104=

I

I0

No, a great earthquake is not twice as intense as a light earthquake, it is 10 000 times as intense as a light earthquake.


a)

n = 1!7 logT

3

n = 1!7 log

1

83

n &= 3.1

The shade number that the welder should use is 3.

b)

n = 1!7 logT

3

14 = 1!7 logT

3

13= !7 logT

3

!39

7= logT

10!

39

7 = T

T &= 2.683"10!6

These glasses transmit a fraction of approximately

1

372 717 of light.


c)

n = 1!7 logT

3

2 = 1!7 logT

3

1= !7 logT

3

!3

7= logT

10!

3

7 = T

T &= 0.373

0.373

2.683!10"6

&= 138 950

The #2 welding glasses transmit approximately 138 950 times as much visible light as the #14 welding glasses.


b) Domain:

T !R, 0 < T "1{ } ; Range:

n !R, n "1{ }

Since T is the fraction of light, it can be between 0 and including 1. T cannot be 0 since log T is undefined. From the graph we can see that see that n is greater than or equal to 1.

Chapter Problem Wrap-Up Chapter Problem Wrap-Up solutions are provided in the 12 Teacher’s Resource.


Chapter 6 Practice Test Chapter 6 Practice Test Question 1 Page 358 The correct solution is D.

Chapter 6 Practice Test Question 2 Page 358 The correct solution is A.

log5

1

125

!"#

$%&= log

5

1

53

!"#

$%&

= log55'3

= '3

Chapter 6 Practice Test Question 3 Page 358 The correct solution is B.

log416!3

= !3log4

42

= !3 2( )= !6

Chapter 6 Practice Test Question 4 Page 358 The correct solution is C. Graph is y = log x translated 3 units to the left and 5 units down. Chapter 6 Practice Test Question 5 Page 359

a)

log6

27 =log 27

log6

&= 1.839

b)

log12

63= 3log

126

=3log6

log12

&= 2.163


Chapter 6 Practice Test Question 6 Page 359

a) g(x) =

1

2log !(x + 3)"# $% + 6

Compress vertically by a factor of

1

2, translate 3 units left, translate 6 units up, and reflect in

the y-axis. b)

c) i) Domain:

x !R, x < "3{ }

ii) Range:

y !R{ } iii) The vertical asymptote occurs when –x – 3 = 0. The vertical asymptote is x = –3. d)

The inverse is an exponential function. Chapter 6 Practice Test Question 7 Page 359



A(t) = P(1+ i)t

2 = (1.065)t

log 2 = t log1.065

t =log 2

log1.065

t &= 11

It will take approximately 11 years for the amount to double. Chapter 6 Practice Test Question 9 Page 359 a)

pH = ! log H+"

#$%

= ! log(3.2 &10!12 )

&= !(!11.5)

&= 11.5

The pH level of ammonia is approximately 11.5. b)

pH = ! log H+"

#$%

10!pH= H

+"#

$%

10!2= H

+"#

$% 10!3.5

= H+"

#$%

H+= 0.01 H +

&= 0.000 32

The range of the hydronium ion concentration in vinegar is approximately 0.000 32 < V < 0.01. c) Vinegar is acidic since its pH is less than 7. Ammonia is alkaline since its pH is greater than 7. Chapter 6 Practice Test Question 10 Page 359 a)


b)

The atmospheric pressure is approximately 54.3 kPa at an altitude of 5 km. c)

The mountain climber is approximately 1.4 km above sea level. d) b)

P = 101.3(1.133)!d

= 101.3(1.133)!5

&= 54.3

c)

P = 101.3(1.133)!d

85 = 101.3(1.133)!d

85

101.3= 1.133!d

log85

101.3

"#$

%&'= !d log1.133

d = !log

85

101.3

"#$

%&'

log1.133

d &= 1.4



a)

!2" !

1= 10 log

I2

I1

#

$%&

'(

!2" 85 = 10 log 315( )!

2= 10 log 315( ) + 85

!2

&= 110

The passing subway train has a level of approximately 110 dB.

b)

!2" !

1= 10 log

I2

I1

#

$%&

'(

118" 85 = 10 logI

2

I1

#

$%&

'(

33= 10 logI

2

I1

#

$%&

'(

3.3= logI

2

I1

#

$%&

'(

103.3=

I2

I1

I2

I1

&= 1995

The sound of the power saw is approximately 1995 times as intense as the sound of normal city traffic.

Chapter 6 Practice Test Question 12 Page 359 Examples may vary. A sample solution is shown. Graphing y = 2log x and y = log x2 results in the same graph when x > 0. The results are the same for y = 6log x and y = log x6 and y = 2.3log x and y = log x2.3.

advanced functions chapter 6 solutions

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